TITLE: Automorphisms of $\mathbb{Z}/p\oplus\cdots\oplus \mathbb{Z}/p$
QUESTION [6 upvotes]: Consider the abelian group
$$G = \underbrace{\mathbb{Z}/p\oplus\cdots\oplus \mathbb{Z}/p}_{n},$$
where $p$ is prime and $1\le n \le p$. I want to show that $G$ has no automorphism of order $p^2$.
I observe that group automorphism of $G$ is the same as linear isomorphism of $G$ as a $\mathbb{Z}/p$-vector space. So the question is the same as to prove that there is no $n\times n$ invertible matrix in $\mathbb{Z}/p$ that has order $p^2$.
My attempt is to calculate the order of $GL(n,\mathbb{Z}/p)$ and show that $p^2$ does not divide that order, but unfortunately this only works for $n\le 2$.

REPLY [2 votes]: A variant on user148212' s argument . Let $g$ be a $p$-element in $G$, that is, $g^{p^{k}} = 1$ for some $p$. Since we are in characteristic $p$, we have $(g - 1)^{p^{k}} = 0$, that is, $g-1$ is nilpotent. Then proceed as in Martin Brandenburg's answer, $(g-1)^{p} = 0$, so that reversing the argument $g^{p} = 1$.